Παρασκευή 25 Οκτωβρίου 2019

HYACINTHOS 27204

[Seiichi Kirikami]:


Dear friends,


The lines X(13)X(15) and X(14)X(16) are parallel to the Euler line of a triangle.

Find the following:

1. the coordinates of the orthogonal projections of X(13), X(14), X(15) and X(16) on the Euler line.

2. the distances of the lines X(13)X(15) and X(14)X(16) from the Euler line.
 
 
[César Lozada]:
 

1)

PX13 = ORTHOGONAL PROJECTION OF X(13) ON EULER LINE

= 2*(a^8-(b^2+c^2)*a^6+b^2*c^2* a^4+(b^2-c^2)^2*b^2*c^2)*sqrt( 3)*S+(a^4+(b^2+c^2)*a^2-2*(b^ 2-c^2)^2)*(a^6-(b^2+c^2)*a^4-( b^4-3*b^2*c^2+c^4)*a^2+(b^4-c^ 4)*(b^2-c^2)) : : (barys)

= on lines: {2,3}, {13, 523}

= [ 0.5098297029399999, -0.3610351635870000, 3.6553058861110000 ]

 

PX14 = ORTHOGONAL PROJECTION OF X(14) ON EULER LINE

= -2*(a^8-(b^2+c^2)*a^6+b^2*c^2* a^4+(b^2-c^2)^2*b^2*c^2)*sqrt( 3)*S+(a^4+(b^2+c^2)*a^2-2*(b^ 2-c^2)^2)*(a^6-(b^2+c^2)*a^4-( b^4-3*b^2*c^2+c^4)*a^2+(b^4-c^ 4)*(b^2-c^2)): : (barys)

= on lines {2,3}

= [ 5.9730124263560000, 5.0877355458610000, -2.6383889389300000 ]

 

PX15 = ORTHOGONAL PROJECTION OF X(15) ON EULER LINE

= 2*(a^8-(b^2+c^2)*a^6+b^2*c^2* a^4+(b^2-c^2)^2*b^2*c^2)*sqrt( 3)*S+3*(a^6-(b^2+c^2)*a^4-(b^ 4-3*b^2*c^2+c^4)*a^2+(b^4-c^4) *(b^2-c^2))*(c^2-a^2+b^2)*a^2 : : (barys)

= on lines: {2, 3}, {15, 523}, {2452, 11485}, {2453, 11480}

= {X(3), X(1316)}-Harmonic conjugate of PX16

= [ 3.5363285717900000, 2.6574797248290000, 0.1687191777380000 ]

 

PX16 = ORTHOGONAL PROJECTION OF X(16) ON EULER LINE

= -2*(a^8-(b^2+c^2)*a^6+b^2*c^2* a^4+(b^2-c^2)^2*b^2*c^2)*sqrt( 3)*S+3*(a^6-(b^2+c^2)*a^4-(b^ 4-3*b^2*c^2+c^4)*a^2+(b^4-c^4) *(b^2-c^2))*(c^2-a^2+b^2)*a^2 : : (barys)

= on lines {2,3}

= {X(3), X(1316)}-Harmonic conjugate of PX15

= [ -5.1846308414960000, -6.0404735773240000, 10.2154373468990000 ]

 

2)

D35 = distance(X(13)X(15), Euler-line) = |(SB-SC)*(SC-SA)*(SA-SB)/(4*S* OH*(SW+sqrt(3)*S))|

D46 = distance(X(14)X(16), Euler-line) = |(SB-SC)*(SC-SA)*(SA-SB)/(4*S* OH*(SW-sqrt(3)*S))|

 

D35/D46 = cos(ω+π/6)/cos(ω-π/6), where ω = Brocard angle of ABC.

 

César Lozada

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